Calc class is covering pythagoras and trig, which is pretty familiar, so I'm slightly bored which leads to mischief.

If I had a right triangle with a known hypotenuse and one angle is it possible to get the lengths of the other two sides without using cos/csc/sin/sec? In other words only using tan/cot and pythagoras?

It seems like it should be possible. With pythagoras I can get the sum of a² and b² and with tan/cot I can get the ratio between a and b. I'm fiddling around on paper trying to do it as a formula but it's not simplifying for me.

With sides a,b,c and angles A,B,C where the matching letters are opposites (side opposite angle) and c is the hypotenuse, given A and c:

a² + b² = c²

(a*b/b)² + (b*a/a)² = c²

(b*a/b)² + (a*b/a)² = c²

(b*tan(A))² + (a*cot(A))² = c²

b²*tan(A)² + a²*cot(A)² = c²

b²*tan(A)² = c² - a²*cot(A)²

b² = (c² - a²*cot(A)²) / tan(A)²

b = sqrt((c² - a²*cot(A)²) / tan(A)²)

So I can solve for b in terms of a or a in terms of b...

Eh... Is this possible? (OMG I'm such a nerd...)

Or maybe something like:

a² + b² = c²

a² = c² - b²

a = sqrt(c² - b²)

tan(A) = a/b

b*tan(A) = a

b*tan(A) = sqrt(c² - b²)

Which eliminates a but I don't know if that b² is recoverable...

Okay, so...

Say A = 36.8699° and c = 10

tan(A) = 3/4

We could say that:

a = 3x

b = 4x

(3x)² + (4x)² = 10²

9x² + 16x² = 100

25x² = 100

x² = 100/25

x² = 4

x = 2

a = 6

b = 8

Which is correct, so this is possible with that method, but now I'm wondering how I can express that as a formula...

Oh, I got it...

Say A = 36.8699° and c = 10

tan(A) = 0.75

a = 0.75b

(0.75b)² + b² = 10²

0.5625b² + b² = 100

1.5625b² = 100

b² = 100/1.5625

b² = 64

b = 8

a = 0.75*8 = 6

Schwing! Achievement unlocked.

(tan(A)² * b²) + b² = c²

(tan(A)² + 1) * b² = c²

b² = c² / (tan(A)² + 1)

b = c / sqrt(tan(A)² + 1)

a = b*tan(A)

Hope someone else enjoys this.

**Edited by Khatharr, 01 February 2013 - 06:18 AM.**